(1-x)/(x^2+1)^2的不定积分如何求,用分式分解为部分分式做,

题目

(1-x)/(x^2+1)^2的不定积分如何求,用分式分解为部分分式做,

答案

没必要用部分分式,直接拆解可以了,前面用三角换元,后面用凑微分
∫ (1 - x)/(x² + 1)² dx
= ∫ dx/(x² + 1)² - ∫ x/(x² + 1)² dx
= ∫ d(tanz)/(tan²z + 1)² - ∫ 1/(x² + 1)² d(x²/2)
= ∫ sec²z/sec⁴z dz - (1/2)∫ 1/(x² + 1)² d(x² + 1)
= ∫ cos²z dz - 1/2 • - 1/(x² + 1)
= (1/2)∫ (1 + cos(2z)) dz + 1/[2(x² + 1)]
= z/2 + (1/4)sin(2z) + 1/[2(x² + 1)] + C
= z/2 + (1/2)sinzcosz + 1/[2(x² + 1)] + C
= (arctanx)/2 + x/[2(x² + 1)] + 1/[2(x² + 1)] + C
= (1/2)arctanx + (x + 1)/[2(x² + 1)] + C
其中tanz = x,dx = sec²z dz
sinz = x/√(x² + 1),cosz = 1/√(x² + 1)